Let $y(t)=s(t)+n(t)$

For the matched filter h(t) on $[0,T]$ we have, $h(t)=s(T-t)$. And the output is: $z(t)=\int_0^T y(u)h(t-u)du=\int_0^T y(u)s(t-(T-u))du=\int_0^T y(u)s(u-(T-t))du$

But sometimes I see, $z(t)=\int_0^T y(u)s(u)du$. How do we get the last equation?

  • 1
    $\begingroup$ $t=T$ the convolution is at its maximum value $\endgroup$ – Stanley Pawlukiewicz May 6 '18 at 13:35

$z(t)$ cannot possibly equal $\displaystyle \int_0^T y(u)s(u) \mathrm du$ unless $z(t)$ has constant value for all $t$: the value of the integral does not depend on $t$ at all! Nor is your convolution integral correct. If $s(t)$ is nonzero only for $t \in [0,T]$, then so is $h(t) = s(T-t)$ nonzero only for $t \in [0,T]$. Consequently, for every real number $t, -\infty < t < \infty$, the matched filter output $z(t)$ is given by \begin{align} z(t) &= \int_{-\infty}^\infty y(u)h(t-u) \,\mathrm du\\ &= \int_{t-T}^t y(u)h(t-u) \,\mathrm du &{h(t-u) = 0~ \text{whenever}~u >t~\text{or}~ u < T-t}\\ &= \int_{t-T}^t y(u)s(u+(T-t)) \,\mathrm du\\ &= \int_{0}^T y(v+T-t)s(v) \,\mathrm dv.\tag{1} \end{align} Now, when $t$ equals $T$, we get that the matched filter output at time $t=T$ is $$z(T) = \int_{0}^T y(v)s(v) \,\mathrm dv = \int_{0}^T y(u)s(u) \,\mathrm du.\tag{2}$$ Note clearly the difference between your $\displaystyle z(t) = \int_0^T y(u)s(u) \mathrm du$ and what you (or your textbook author) should have written, namely $(2)$.

As @Stanley Pawlukiewicz points out, the matched filter output has a maximum value at $t=T$. For this result, (and far more than you probably want to know about matched filters), see this answer of mine on this site.


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